Answer:
Work done on an object is equal to
FDcos(angle).
So, naturally, if you lift a book from the floor on top of the table you do work on it since you are applying a force through a distance.
However, I often see the example of carrying a book through a horizontal distance is not work. The reasoning given is this: The force you apply is in the vertical distance, countering gravity and thus not in the direction of motion.
But surely you must be applying a force (and thus work) in the horizontal direction as the book would stop due to air friction if not for your fingers?
Is applying a force through a distance only work if causes an acceleration? That wouldn't make sense in my mind. If you are dragging a sled through snow, you are still doing work on it, since the force is in the direction of motion. This goes even if velocity is constant due to friction.
Explanation:
See coulomb's law. Force is inversely proportional to the distance squared. So if you multiply r by 2, the force is multiplied by (½)² = ¼.
a. F/4
Answer:
The transmitted intensity through all polarizers is 
Explanation:
According to Malu's law the intensity of a polarized light having an initial intensity 
is mathematically represented as
Now considering the polarizer(The polarizing disk) the equation above becomes
Where n is the number of polarizers
Substituting 
for the initial intensity 3 for the n and 20° for the angle of rotation
Answer:
Hi myself Shrushtee.
Explanation:
The fuse is connected to the live wire so that the appliance will not become charged (have a potential difference of 230 V) after the fuse has melted due to excessive current. Fuses must be fitted onto the live wire so that when it blows, it will disconnect (isolate) the appliance from the high voltage live wire.
Answer:
I should be active for 15 hours to meet the physical activity requirement.
Explanation:
Since time dilates in moving objects, we use the formula t = t₀/√(1 - β²) where t = time in space vehicle, t₀ = time on earth = 9 hours and β = v/c where v = speed of space vehicle = 0.8c.
So, t = t₀/√(1 - β²)
t = 9/√(1 - (v/c)²)
= 9/√(1 - (0.8c/c)²)
= 9/√(1 - (0.8)²)
= 9/√(1 - (0.64)
= 9/√0.36
= 9/0.6
= 15 hr
So, according to a timer on the space vehicle, I should be active for 15 hours to meet the physical activity requirement.
